The homology groups of $S^n\vee S^m$.

Question

I used the Mayer-Vietoris to compute the homology group of $S^n \vee S^m$. When $q=n,q\neq m$,$H_q(S^n \vee S^m)=\mathbb{Z}$, but the outcome is quite different from the thorem 2.6 which was presented in Fermin Dalmagroi's paper, see the link: http://www.scielo.org.co/pdf/rcm/v39n1/v39n1a03.pdf

Answer 1

I guess you're talking about their Theorem 2.6 which reads:

Theorem 2.6. Given $m,n \in \mathbb{N}$, let us consider $S^n \vee S^m$, the wedge sum of spheres $S^n$ and $S^m$. For each $q > 0$ we have $$H_q(S^n \vee S^m) = \begin{cases} \mathbb{Z} \oplus \mathbb{Z} & \text{if } q = n = m \\ 0 & \text{in any other case.} \end{cases}$$

To be honest I'm not sure how to understand this theorem, because in the most straightforward way I can think of, it is wrong... Here is what is correct (maybe there's a way to interpret their theorem in a way that matches this, the phrasing is a bit weird, but since after that they consider only the case $n=m$ it may just be a small oversight), for simplicity I assume $n,m > 0$:

• if $n = m$, then $$H_q(S^n \vee S^n) = \begin{cases} \mathbb{Z} & q = 0 \\ \mathbb{Z} \oplus \mathbb{Z} & \text{if } q = n = m \\ 0 & \text{otherwise.} \end{cases}$$

• if $n \neq m$, then $$H_q(S^n \vee S^m) = \begin{cases} \mathbb{Z} & \text{if } q \in \{0,n,m\} \\ 0 & \text{otherwise.} \end{cases}$$